Question

In figura 2 este reprezentat un trapez isoscel ABCD cu AB || CD, AC si BD
perpendiculare, AB=8 cm, CD=4 cm. Punctele M.N.P şi Q sunt mijloacele
laturilor AB, BC, CD respectiv DA si O este punctul de intersectie a
diagonalelor trapezului.
1. Lungimea segmentului AD este egală cu....
2. Aria patrulaterului MNPQ este egala cu....​

Answer 1

 

$\displaystyle\bf\\Se~da:\\\\Trapezul~isoscel~ABCD~cu~AB||CD\\AB=8~cm\\CD=4~cm\\AC\perp BD\\AC\cap BD=\{O\}\\Punctele~M,N,P,~respectiv~Q~sunt~mijloacele\\laturilor~AB,BC,CD~respectiv~DA\\\\Se~cere:\\\\1)~~AD=~?\\2)~~Aria~patrulaterului~MNPQ=~?$

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$\displaystyle\bf\\Rezolvare~1):~(Vezi~Figura~1~atasata.)\\\\QN~este~linie~mijlocie~in~trapezul~ABCD\\QN=\frac{AB+CD}{2}=\frac{8+4}{2}=\frac{12}{2}=6~cm\\Diagonalele~trapezului~sunt~perpendiculare.\\\\DE=QN=6~cm~deoarece~trapezul~este~ortodiagonal\\\\AE=BF=\frac{AB-CD}{2}=\frac{8-4}{2}=\frac{4}{2}=2~cm\\\\In~\Delta ADE~avem:\\DE=6~cm=cateta\\AE=2~cm=cateta\\AD=?~cm=ipotenuza\\\\AD=\sqrt{DE^2+AE^2}=\sqrt{6^2+2^2}=\\\\=\sqrt{36+4}=\sqrt{40}=\sqrt{4\times10}=\boxed{\bf2\sqrt{10}~cm}$

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$\displaystyle\bf\\Rezolvare~2)~(Vezi Figura~2~atasata)\\\\In~\Delta~ADC,~PQ~este~linie~mijlocie.\\\implies~PQ||AC~si~PQ=\frac{AC}{2}\\\\In~\Delta~ABC,~MN~este~linie~mijlocie.\\\implies~MN||AC~si~MN=\frac{AC}{2}\\\\\implies\implies~\boxed{\bf PQ||MN~si~PQ=MN}\\\\\\In~\Delta~BAQ,~MQ~este~linie~mijlocie.\\\implies~MQ||BD~si~MQ=\frac{BD}{2}\\\\In~\Delta~BCQ,~NP~este~linie~mijlocie.\\\implies~NP||BD~si~NP=\frac{BD}{2}\\\\\implies\implies~\boxed{\bf MQ||NP~si~MQ=NP}$

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$\displaystyle\bf\\Rezulta~ca~patrulaterul~MNPQ~are~laturile\\paralele~si~egale~2~cate~2.\\\\Asta~este~d efinitia~paralelogramului.\\\implies~MNPQ~este~paralelogram~(intr-o~prima~aproximare.)\\\\Diagonalele AC=BD~deoarece~ABCD~este~trapez~isoscel.\\\\\implies~Paralelogramul~MNPQ~are~toate~laturile~egale.\\\\\implies~MNPQ~este~romb~(caz~particular~de~paralelogram).$

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$\displaystyle\bf\\MP=inaltimea~trapezului\\Am~stabilit~mai~sus~ca~inaltimea = linia~mijlocie\\MP=NQ~=6~cm\\\\Aria~rombului~este~produsul~diagonalelor~supra~2.\\\\Aria~MNPQ=\frac{MP\times NQ}{2}=\frac{6\times6}{2}=\frac{36}{2}=\boxed{\bf18~cm^2}\\\\Diagonalele~trapezului~sunt~perpendiculare.\\\\AC\perp BD\\\implies~MNPQ~este~patrat~(caz~particular~de~romb.)\\Formula~pentru~arie~folosita~la~romb~ramane~valabila~si~la~patrat.$